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Amp Meter

Discussion in 'Electronic Basics' started by Nikki, Feb 15, 2005.

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  1. Nikki

    Nikki Guest

    Hi guys
    I have a circuit with a 24 volt DC power supply 2 amps that feeds about 20
    small relays and I have an amp meter that just says 1 to 5 on it. How would
    I put this in the circuit I think it would go in series but to draw 2 amps
    through the meter would that not damage it
    Thanks
    Nikki
     
  2. Jamie

    Jamie Guest

    most likely the meter is already designed to be used inline directly.
    other wise you would need to know the R of the meter coil and voltage
    of the meter to calculate for a shunt resistor.
     
  3. Rich Grise

    Rich Grise Guest

    The meter itself should tell you, if the face is still legible. Since it's
    a 1-5 amp meter, it should be OK, but if it has fine print near the bottom
    of the face, read that, and if that doesn't tell you exactly how to
    connect it, then post what it says, and we'll try to answer better.

    Just in case I'm missing something, it _does_ have a 0 to the left of
    the 1, doesn't it?

    If so, and if there's nothing in fine print, and nothing stamped or molded
    into the back of the meter, and if the studs are pretty good-sized, (like
    5 mm dia. or fatter), then it's probably OK to just put it in series.
    First thing out of the supply, of course. ;-)

    Good Luck!
    Rich
     
  4. Nikki

    Nikki Guest

    On closer inspection of the meter I have taken the cover off and the needle
    will not move so I think the meter is no good. I have a second I would like
    to use this one goes from 0 to 60 On the back it has a + on one of the posts
    and printed on the back is 120 on the front of the meter is says (its very
    small print) FS-50mvDC and it also says D C Amperes. Now I know this came
    from the Phone company and it was on a 48 volt circuit.
    Nikki
     
  5. It sounds like the meter mechanism inside is a 50mV full scale
    instrument, but it is very likely that it has a series or parallell
    resistor inside the cover to make it suitable for 48 Volt or a few
    amperes.

    Open the cover and try to find a series or parallell resistor. Or simply
    test the meter very carefully, using a battery and a high value resistor
    in series, to limit the current to less than 10mA to begin with. If there
    is no reaction change to a ten times lower resistor until it works.
     
  6. Amp meters have a sensitive movement connected in parallel with a low
    resistance current carrying shunt that produces the small voltage
    required to move the meter (50 mV in this case). The open question
    with any amp meter is whether that low resistance shunt is sealed
    inside the meter or is assumed to be external to the meter. Measure
    the resistance of your amp meter. If it is less than an ohm, the
    shunt is almost certainly inside the meter, and it is safe to pass
    amperes through the meter terminals. If higher than an ohm, the meter
    is supposed to have an external shunt resistor connected to it, and
    the load current is supposed to be passed through that resistor
    through an extra pair of terminals. Here is what an external shunt
    resistor might look like:
    http://www.carrel.co.nz/shunt/main.htm
    Note the big terminals for the load current and the smaller ones for
    the meter connection.

    A more detailed review of ammeters:
    http://www.allaboutcircuits.com/vol_1/chpt_8/4.html
     
  7. Rich Grise

    Rich Grise Guest

    DO NOT DO THIS WITH AN OHMMETER!!!

    Use two potentiometers.

    Start with one in series with the meter and, say, a 1.5V battery. Start
    with the pot at max. resistance. Turn it down until the meter reads
    full-scale. Start with at least 10K. If a 1.5V battery won't give full-
    scale deflection even with the pot turned to 0, then clearly the meter
    expects the full 60A through itself, and you'd stick it in series with
    the supply. Otherwise, go to step 2:

    2: Take the second pot and put it in parallel with the meter (while
    leaving the pot and battery, that have your meter deflected full-scale).
    Adjust the second pot until the meter deflects exactly one-half. Take that
    resistor, and measure _its_ resistance with the ohmmeter, and that will
    tell you the internal resistance of the meter.

    Then, it's just Ohm's law. :)

    Good Luck!
    Rich
     
  8. Jamie

    Jamie Guest

    you will need to determine the coil R (resistance), 120 maybe it.
    lets assume for the moment that it is.
    your scale is 0..60, we will use that for amp scale.

    Imeter = 0.050/120 = 416 uA ( does not seem likely etc).

    Rshunt = 0.050/(60.0 - 0.000416) = 0.00083 ohms

    so this would mean you need a shunt resistor on the
    meter terminals of 0.00083 which is more like a bar
    of copper across the terminals of the meter.
    if the 120 number you are reading is actually indicating
    I of the meter then you use 120 uA or 120 mA instead of the
    416 uA in the first calculation.

    in either case, you will need to refer to a wire&meteral
    chart to get teh ohms per 1K' (per 1k feet) to determine what
    you need to create a shunt .

    these are only my thoughts off the top of my head going by the
    info your passing.
     
  9. Nikki

    Nikki Guest

    Guys thanks for all your replies
    I took the suggestion of Rich and found that with a 1.5volt battery and a
    resistance of 10400 ohms I can get the meter to read at its last increment.
    The meter is marked in increments of 10 up to 60 how would I get a 2 amp
    draw to read 20 where a 6 amp draw would be the last increment on the meter
    Thanks
    Nikki
     
  10. Let's imagine a circuit in our minds. The instrument is in parallell with
    a big shunt, we are seeing it deflect to the full 60 Amp, as you wanted
    it.

    What do we know about this circuit?

    The instrument is a 65mV 148uA full scale instrument.
    (1.5V/10400 Ohm = 148uA, and wasn't it a 65mV instrument?)

    The voltage over the shunt must be 65mV too.

    This shunt must have a resistance of 65mV divided by 20 Amps, which is
    0.00325 Ohm.

    So you need to find a piece of metal with this resistance from end to
    end, then connect your instrument in parallell to it, and you have a 60
    Amp current meter.

    Make it a little longer than needed, and adjust the connection points of
    the meter to the shunt to adjust it. Connect the outer current at the
    real ends of the shunt.

    That would create an instrument which does not need a power supply, no
    battery.

    If we add a battery we can choose a shunt much more freely, as we can use
    an opamp to scale the result to fit to the instrument. It will give
    adjustment possibilities and it is a lot easier to find a shunt.
     

  11. I see now that you wanted a full scale of 6 Amp.
    Well, adjust the calculations accordingly.
     
  12. There is also a practical way to solve this problem without calculations.

    Try different pieces of metal, pieces of rebar, metal bars of all kinds.

    Send a known current through it. Put the probes from the instrument
    together at the middle of the metal bar. Pull the probe tips apart slowly
    and watch how the instrument shows higher indications. When at the full
    scale, or the number you want, mark the two points.

    Cut off the metal bar right outside the marks.
    Connect the probes at their marked points, and the outer current from the
    outer ends of the metal bar.
     
  13. Don Kelly

    Don Kelly Guest

    ---------------
    You will have to do a bit better. If you have a voltmeter, measure the
    voltage at the battery- it is unlikely to be 1.5V. You could also measure
    the voltage across the potentiometer and determine the actual current. It
    appears to be a 50mv meter which seems to be reading full scale at
    120ma.giving a resistance of 0.417 ohms.
    If this is so -then for 6A full scale, the meter will need an external shunt
    of 0.0085 ohms. You will need a wire large enough to carry 6A and long
    enough to get 0.0085 ohms (4 inches of #24 wire may do).
    However- check the actual current and voltage as above as the FS reading may
    not be 120ma.
    I assume you are not looking for great accuracy.
    See calculations below
    V=meter FS voltage. Im=meter FS current. I=desired full scale current
    Rshunt =V/(I-Im)
    In your case V=50mv, I=6A and, if the assumption above is correct, Im
    =0.120A
    then Rshunt =0.05/(6-0.12) =0.0085 ohms
    #24 wire should carry 6A and has a resistance at 20degrees C of 25.7
    ohms/1000ft so 4 inches should do.

    Someone please check the above.
     
  14. Nikki

    Nikki Guest

    Thanks guys
    If I have a circuit lets say 1.5 volt bat and a 1.5 volt lamp and I want to
    measure the current I would put the meter parallel to a shunt resistor in
    the circuit right. But does the shunt resistor not cut down on the voltage
    to the lamp
     
  15. I think you are right.

    In practical terms: Take a piece of copper wire, diameter 1-2mm.
    Send 1 Amp through it. Put the wires from the meter together on that
    wire. Pull them slowly apart until the meter reads 1 Amp. Mark these
    measuring points and solder the instrument wires to them. Cut off the
    thick wire outside the measuring marks and attach the current connections
    to the ends of the wire.
     
  16. Rich Grise

    Rich Grise Guest

    I've also seen a circuit where a microammeter is turned into a voltmeter -
    in this case, 10.4K gives 1.5V full-scale. A shunt resistor that drops
    1.5V at 60A would be .025 ohm, which is much more manageable of a value
    than 0.0083 ohm, and the calibration isn't anywhere near as persnickety.

    Good Luck!
    Rich
     
  17. Rich Grise

    Rich Grise Guest

    Yes, by 0.05 volts. :)

    But don't just slap a shunt in there and expect to get accurate readings.
    Either give yourself a way of trimming it while live, or actually
    determine the internal resistance of the meter. Take your setup with
    the batt. and series R, adjust it for full-scale, as you've done, and
    then take another pot, put it in parallel with the meter (so the meter
    and new pot are now in parallel with each other, and that combination
    is in series with the B and R), and adjust it until the meter reads
    half-scale.

    But this will only work if the meter actually _has_ a Zero indication.

    Did you meant to say that your meter face has "1" at the very leftmost
    spot? If so, you might have to do some more experiments to find out
    exactly what this meter is expecting - usually, when there's a meter
    that doesn't start at 0, it's used with some kind of custom circuit to
    get that offset.

    Of course, the other thing (depending on how much headroom you have)
    is to turn the meter into a voltmeter, by just adding a series resistor,
    and measuring the voltage across the shunt. Doing that gets your
    calibration up out of the mud. :)

    Good Luck!
    Rich
     
  18. In this case you are wasting a major part of the signal. That is not a
    good idea, seen from an engineering point of view.

    We want a shunt which gives as low voltage drop as possible. Your
    idea would make the voltage drop many times bigger, from 50mV to 1500mV.

    The 0.0083 Ohm shunt is easy to construct from heavy gauge copper wire.
    Or a thinner wire laid double, in quadruple, etc..

    Adjustment can be made by moving the probe measuring point a little, or
    by connecting an adjustment pot in parallell with the shunt. A wirewound
    pot will be suitable for this.
     
  19. Jamie

    Jamie Guest

    that looks to be a 150 uA meter at full scale.
    if memory serves,.i think you said the meter was a 0.050 mv?

    to get a 6 amp scale.
    6.00 - .000150 = 5.999850

    Shunt = 0.050 / 5.999850 = 0.0083 ohms

    i think that is correct,i normally measure the
    coil R and use the I (current) to do all of this
    how ever, this should still work.
    this is just off the top of my head, i am sure some
    one will correct me.
     
  20. Rich Grise

    Rich Grise Guest

    Well, in my defense, in a different post, I did mention that this
    technique depends on the headroom of the supply. And you can certainly
    decrease the ranging resistors, to decrease the shunt drop. It's just
    a seat-of-the-pants sort of thing - you decrease the pickiness of the
    shunt by adding _some_ resistance in series with the meter.

    I've calibrated ammeter shunts, and it's a PITA. They wouldn't let
    me grind on the brass strip with the Dremel while power was on. ;-)

    I've also seen ammeters that have no leads at all - they have a clip
    on the back that clips over the cable, and the meter itself is nothing
    but a spring-loaded compass. You calibrate them by moving the cable
    around until the meter reads right, then hot-glue it.

    Cheers!
    Rich
     
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